Let \(T_j\)  and \(U_j\)  denote the Chebychev polynomials
of the first and second kind, respectively. Keeping in mind
the representation \(T_j(\cos (\theta ))=\cos (j\theta )\)  and
\(U_{j-1}(\cos (\theta ))=\sin (j\theta )/\sin (\theta )\) 
([1]}) from the conditions
(REF )-() we get
\(A_j:=\frac{1}{\pi }\int _{-\pi }^{\pi }\widetilde{\Phi }^a(e^{i\theta })\cos (j\theta )d\theta =0,\quad j=0,1,...,s-1\\B_j:=\frac{1}{\pi }\int _{-\pi }^{\pi }\widetilde{\Phi }^a(e^{i\theta })\sin (j\theta )d\theta =0,\quad j=1,...,s-1.\) 
